# Trigonometric Limits

The basic trigonometric limit is

$\lim\limits_{x \to 0} \frac{{\sin x}}{x} = 1.$

Using this limit, one can get the series of other trigonometric limits:

$\lim\limits_{x \to 0} \frac{{\tan x}}{x} = 1,\;\;\; \lim\limits_{x \to 0} \frac{{\arcsin x}}{x} = 1,\;\;\; \lim\limits_{x \to 0} \frac{{\arctan x}}{x} = 1.$

Further we assume that angles are measured in radians.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the limit $\lim\limits_{x \to 0} {\frac{{4x}}{{\sin 3x}}}.$

### Example 2

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\cos {3x} - \cos x}}{{{x^2}}}}.$

### Example 3

Find the limit $\lim\limits_{x \to 0} {\frac{{\sin5x - \sin 3x}}{{\sin x}}}.$

### Example 4

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 0} {\frac{{4x}}{{\sin 3x}}}.$

Solution.

$L = \lim\limits_{x \to 0} \frac{{4x}}{{\sin 3x}} = \lim\limits_{x \to 0} \frac{{3 \cdot 4x}}{{3\sin 3x}} = \frac{4}{3}\lim\limits_{x \to 0} \frac{{3x}}{{\sin 3x}} = \frac{4}{3}\lim\limits_{x \to 0} \frac{1}{{\frac{{\sin 3x}}{{3x}}}} = \frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}}.$

Since $$3x \to 0$$ as $$x \to 0,$$ we can write:

$L = \frac{4}{3}\frac{{\lim\limits_{x \to 0} 1}}{{\lim\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}} = \frac{4}{{3\lim\limits_{3x \to 0} \frac{{\sin 3x}}{{3x}}}} = \frac{4}{{3 \cdot 1}} = \frac{4}{3}.$

### Example 2.

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\cos {3x} - \cos x}}{{{x^2}}}}.$

Solution.

We factor the numerator:

$\cos{3x} - \cos x = - 2\sin \frac{{3x - x}}{2}\sin \frac{{3x + x}}{2} = - 2\sin x\sin 2x.$

This yields

$\lim\limits_{x \to 0} \frac{{\cos 3x - \cos x}}{{{x^2}}} = \lim\limits_{x \to 0} \frac{{\left( { - 2\sin x\sin 2x} \right)}}{{{x^2}}} = - 2\lim\limits_{x \to 0} \frac{{\sin x}}{x} \cdot \lim\limits_{x \to 0} \frac{{\sin 2x}}{x} = - 2 \cdot 1 \cdot \lim\limits_{2x \to 0} \frac{{2\sin 2x}}{{2x}} = - 2 \cdot 2\lim\limits_{2x \to 0} \frac{{\sin 2x}}{{2x}} = - 4.$

### Example 3.

Find the limit $\lim\limits_{x \to 0} {\frac{{\sin5x - \sin 3x}}{{\sin x}}}.$

Solution.

We use the following trigonometric identity:

$\sin x - \sin y = 2\sin \frac{{x - y}}{2}\cos \frac{{x + y}}{2}.$

Then we obtain

$\lim\limits_{x \to 0} \frac{{\sin5x - \sin 3x}}{{\sin x}} = \lim\limits_{x \to 0} \frac{{2\sin \frac{{5x - 3x}}{2} \cos \frac{{5x + 3x}}{2}}}{{\sin x}} = \lim\limits_{x \to 0} \frac{{2\sin x\cos 4x}}{{\sin x}} = \lim\limits_{x \to 0} \left( {2\cos 4x} \right).$

As $$\cos{4x}$$ is a continuous function at $$x = 0,$$ then

$\lim\limits_{x \to 0} \left( {2\cos 4x} \right) = 2\lim\limits_{x \to 0} \cos 4x = 2 \cdot \cos \left( {4 \cdot 0} \right) = 2 \cdot 1 = 2.$

### Example 4.

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x}}.$

Solution.

Using the trig identity

$\cos \alpha - \cos \beta = - 2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2},$

we convert the limit in the following way:

$L = \lim\limits_{x \to 0} \frac{{\cos \left( {x + a} \right) - \cos \left( {x - a} \right)}}{x} = - 2\lim\limits_{x \to 0} \frac{{\sin x\sin a}}{x}.$

Here $$\sin a$$ is a constant that does not depend on $$x.$$ Therefore,

$L = - 2\sin a\,\underbrace {\lim\limits_{x \to 0} \frac{{\sin x}}{x}}_1 = - 2\sin a.$